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$$\int_{0}^1 (1 + 4y^2)^{1/2} dy$$

So, how do I integrate this without the use of trigonometrical substitution? Can anybody give me a hint? Thank you!

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  • $\begingroup$ Try $y=\frac{\sinh\theta}{2}=\frac{e^\theta-e^{-\theta}}{4}.$ $\endgroup$ Commented Apr 26, 2015 at 17:20
  • $\begingroup$ COuld you develop where you got this ??? $\endgroup$ Commented Apr 26, 2015 at 17:21
  • $\begingroup$ @JackD'Aurizio You beat me to it by a few seconds, though of course one might object that this is cheating w.r.t. the prohibitions on trigonometric functions! $\endgroup$ Commented Apr 26, 2015 at 17:22
  • $\begingroup$ Given the hyperbolic Pythagoran theorem $\cosh^2 = 1+\sinh^2$ we have that such a substitution simplifies the integral. $\endgroup$ Commented Apr 26, 2015 at 17:22
  • $\begingroup$ @Travis: the primitive is related with the $\operatorname{arcsinh}$ function so there is no way of solving the integral without trigonometrical or hyperbolic trigonometrical (i.e. exponential) substitutions. $\endgroup$ Commented Apr 26, 2015 at 17:23

1 Answer 1

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The idea is to use the identity $$ \cosh^2 u = \sinh^2 u + 1 $$

to simplify the square root; so, let $$ \sinh^2 u = 4y^2 \to \sinh u = 2y \to u = \sinh^{-1} 2y \\ \implies \cosh u du = 2dy \\ \sqrt {1 + 4y^2} = \cosh u \\ $$ from this you should be able to reoslve the integral.

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